Problem 46 - College Trigonometry

Question

For the given vector \(\vec{v}\), find the magnitude \(\|\vec{v}\|\) and an angle \(\theta\) with \(0 \leq \theta<360^{\circ}\) so that \(\vec{v}=\|\vec{v}\|\langle\cos (\theta), \sin (\theta)\rangle\) (See Definition 11.8.) Round approximations to two decimal places. $$ \vec{v}=\langle-2,-1\rangle $$

Step-by-Step Solution

Answer

The magnitude of \( \vec{v} \) is approximately 2.24, and angle \( \theta \) is 116.57°.

Step 1: Calculate Magnitude

To find the magnitude of vector \( \vec{v} = \langle -2, -1 \rangle \), use the formula \( \| \vec{v} \| = \sqrt{x^2 + y^2} \). Substitute \( x = -2 \) and \( y = -1 \) into the formula: \[ \| \vec{v} \| = \sqrt{(-2)^2 + (-1)^2} = \sqrt{4 + 1} = \sqrt{5}. \] The magnitude \( \| \vec{v} \| \) is approximately 2.24.

Step 2: Find Angle with Cosine

We know \( \cos(\theta) = \frac{x}{\| \vec{v} \|} \). Substitute \( x = -2 \) and magnitude \( \| \vec{v} \| = \sqrt{5} \): \[ \cos(\theta) = \frac{-2}{\sqrt{5}}. \] Using a calculator (ensure it is set to degrees), find \( \theta \) such that \( \cos(\theta) = -0.894427 \). This gives \( \theta = 116.57^{\circ} \).

Step 3: Find Angle with Sine

We know \( \sin(\theta) = \frac{y}{\| \vec{v} \|} \). Substitute \( y = -1 \) and magnitude \( \| \vec{v} \| = \sqrt{5} \): \[ \sin(\theta) = \frac{-1}{\sqrt{5}}. \] Using a calculator, find \( \theta \) such that \( \sin(\theta) = -0.447214. \) Verify \( \theta \). Here \( \theta \) is also \( 116.57^{\circ} \), confirming earlier calculation.

Step 4: Summary of Angle

Given both calculations of \( \cos(\theta) \) and \( \sin(\theta) \) support \( \theta = 116.57^{\circ} \). Ensure that this angle satisfies the position of the vector in the quadrant coordinate plane, which it does as \( (-2, -1) \) is in the third quadrant where both sine and cosine are negative.

Key Concepts

Understanding Vector Magnitude

Vectors are fundamental components in physics and engineering, crucial for representing quantities with both magnitude and direction. The magnitude of a vector essentially tells you how "long" the vector is in a geometric sense. It is calculated using the formula: \[ \| \vec{v} \| = \sqrt{x^2 + y^2} \]\where \(x\) and \(y\) are the components of the vector. For example, for the vector \( \vec{v} = \langle -2, -1 \rangle \), the magnitude is calculated as follows:

• First, square each of the vector's components: \((-2)^2 = 4\) and \((-1)^2 = 1\).
• Next, sum these squares: \(4 + 1 = 5\).
• Finally, take the square root of this sum to get the magnitude: \(\sqrt{5} \approx 2.24\).

The magnitude is important because it gives us a scalar measure of the vector, which can be used to normalize the vector or find its direction in space.

Exploring the Vector Angle

The angle of a vector is crucial in determining the direction of the vector in a coordinate system. When finding the angle \(\theta\) that a vector makes with the horizontal axis, both the cosine and sine functions are utilized. Given a vector \(\vec{v} = \langle x, y \rangle\), this angle helps us express the vector in its trigonometric form: \(\vec{v} = \|\vec{v}\| \langle \cos(\theta), \sin(\theta) \rangle\). For our vector \(\langle -2, -1 \rangle\):

• Calculate \(\cos(\theta) = \frac{x}{\|\vec{v}\|} = \frac{-2}{\sqrt{5}}\).
• Calculate \(\sin(\theta) = \frac{y}{\|\vec{v}\|} = \frac{-1}{\sqrt{5}}\).
• Both calculations confirm an angle \(\theta \approx 116.57^\circ\).

This angle \(116.57^\circ\) indicates the vector's direction relative to the positive x-axis, confirming its position in the third quadrant. Understanding the vector's angle is key to transitioning between vector calculations and geometric intuition.

The Role of Cosine and Sine Functions

Cosine and sine functions are foundational in trigonometry, particularly when working with vectors. They help to define the direction a vector points by translating angular information into a format that's usable in calculations. For a vector \(\vec{v}\), the expressions \(\cos(\theta)\) and \(\sin(\theta)\) are obtained as follows:

• \(\cos(\theta)\) represents the ratio of the vector's x-component to its magnitude, expressed as \(\cos(\theta) = \frac{x}{\|\vec{v}\|}\).
• Similarly, \(\sin(\theta)\) represents the ratio of the vector's y-component to its magnitude, expressed as \(\sin(\theta) = \frac{y}{\|\vec{v}\|}\).

These functions are valuable not only because they allow the conversion of the vector into its trigonometric form, making it easier to solve problems related to angles and make angle analyses, but also because they confirm the geometric relationships inherent in the vector's placement in the coordinate system. Understanding how these functions relate to a vector's components helps in visualizing and solving complex trigonometric problems efficiently.